Question

The maximum vertical height reached by a projectile is one fourth of the horizontal range,
If the horizontal range is 50 m, find the velocity of projection and the time taken by the
projectile to reach to the maximum height. (g = 9.8 m/s2) [Ans : v= 22.13 m/s, t= 1.59 s​

Answer 1

Answer:

Explanation:

H=R/4    H=25/2

ACC TO FORMULA DERIVED      4H=Rtan∅

tan∅=1  ∅=45

R=u²sin2∅/g

50x10/sin2(45)=u²

u=10√5=22.13m/s

Time of ascent =